Problem: Multiply the following complex numbers: $({1+5i}) \cdot ({-3-i})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({1+5i}) \cdot ({-3-i}) = $ $ ({1} \cdot {-3}) + ({1} \cdot {-1}i) + ({5}i \cdot {-3}) + ({5}i \cdot {-1}i) $ Then simplify the terms: $ (-3) + (-1i) + (-15i) + (-5 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ -3 + (-1 - 15)i - 5i^2 $ After we plug in $i^2 = -1$ , the result becomes $ -3 + (-1 - 15)i - (-5) $ The result is simplified: $ (-3 + 5) + (-16i) = 2-16i $